Question

Rezolvati in R^2 sistemele:
2x+3y=1
X^2+y^2=2

Answer 1

Explicație pas cu pas:

Scoatem x din prima ecuatie:

$2x+3y=1\\2x=1-3y\\x=\frac{1-3y}{2}$

Introducem aceasta relatie in a doua ecuatie a sistemului:

$(\frac{1-3y}{2})^2+y^2=2\\\frac{(1-3y)^2}{2^2}+y^2=2\\\frac{1-6y+9y^2}{4}+y^2=2|*4\\1-6y+9y^2+4y^2=8\\13y^2-6y-7=0\\\Delta=(-6)^2-4*13*(-7)=400\\y_1=\frac{6+20}{26}=1\\y_2=\frac{6-20}{26}=\frac{-14}{26}=\frac{-7}{13}$

Daca $y=1$, atunci $=\frac{1-3*1}{2}=-1$.

Daca $y=\frac{-7}{13}$, atunci $x=\frac{1-3y}{2}=\frac{1-3\frac{-7}{13}}{2}=\frac{1+\frac{21}{13}}{2}=\frac{\frac{13+21}{13}}{2}=\frac{34}{26}=\frac{17}{13}$.